AP EAMCET201921 Apr 2019Morning ShiftMathematicsIndefinite IntegrationActual
In I_n= n x x d x for n=1,2,3, , then I₆=
Options
- A3 5 3 x+ 8 5 ^5 x- x+c
- B2 5 5 x- 5 3 ^3 x-2 x+c
- C2 3 5 x- 8 3 ^5 x+4 x+c
- D2 5 5 x- 8 5 ^3 x+4 x+c
Correct answer
D. 2 5 5 x- 8 5 ^3 x+4 x+c
Step-by-step solution
Subtracting Eq. (ii) from Eq. (i), we get aligned & I_n-I_ n-2 = n x- (n-2) x x d x & = 2 (n-1) x x x d x= 2 (n-1) x d x & = 2 (n-1) x (n-1) aligned I₆-I₄= 2 5 x 5 and I₄-I₂= 2 3 x 3 Now, I₂= 2 x x d x = 2 x x x d x=2 x d x =2 x+c and aligned I₆ & =I₄+2 5 x 5 & =I₂+2 3 x 3 +2 5 x 5 aligned aligned & =2 5 x 5 +2 3 x 3 +2 x+c & = 2 5 x 5 + 2 3 (3 x-4 ^3 x )+2 x+c & I₆= 2 5 5 x- 8 3 ^3 x+4 x+c aligned